Answer to Question #170538 in Real Analysis for Prathibha Rose

Question

Let m,n be non negative integers and.let i, j element of N be even . let f:R² to R be defined by f(0,0)=0 and

f(x,y) = (x^myⁿ)/(xⁱ+y^j) for (x,y) not equal to (0,0) .show that f is continuous at (0,0) if and only if ,if m_j+m_i >i_j

Answer 1

Consider the path $x(t)=t^j, y(t)=kt^i$, which approaches to (0,0) as t tends to 0.

If f(x,t) is a continuous at (0,0) then $f(x(t),y(t))=(1+k^j)^{-1}t^{mj+ni-ij}$ must have a bounded limit as t tends to 0. Hence it must be $mj+ni-ij\geq 0$ . and then

$\lim\limits_{t\to 0}f(x(t),y(t))=\begin{Bmatrix} 0 & if\, mj+ni-ij> 0\\ (1+k^j)^{-1} & if\, mj+ni-ij= 0 \end{Bmatrix}$

So, if $mj+ni-ij= 0$ then the limit value varies depending on k, hence, $\lim\limits_{(x,y)\to (0,0)}f(x,y)$ doesn't exist and the function may not be continuous at (0,0).

So, the necessity of condition $mj+ni-ij> 0$ or, equivalently, $\frac{m}{i}+\frac{n}{j}>1$ is proved.

Let us show now that this condition is sufficient for the function f(x,y) to be continuous at (0,0).

Let $s=\frac{m}{i}+\frac{n}{j}$, $p=\frac{is}{m}$, $q=\frac{js}{n}.$ Then $\frac{1}{p}+\frac{1}{q}=1$.

Since i and j are even positive integers, $x^i+y^j=|x|^i+|y|^j$ for every x,y.

By Young's inequality for products we have

$|x|^i+|y|^j=(|x|^{m/s})^p+(|y|^{n/s})^q=\frac{(p^{1/p}|x|^{m/s})^p}{p}+\frac{(q^{1/q}|y|^{n/s})^q}{q}\geq p^{1/p}q^{1/q}|x|^{m/s}|y|^{n/s}=C(|x|^m |y|^n)^{1/s}$

where $C=p^{1/p}q^{1/q}$. So,

$|f(x,y)|=\frac{|x|^m|y|^n}{|x|^i+|y|^j}\leq C^{-1}(|x|^m|y|^n)^{1-1/s}\to 0$

Therefore, if $mj+ni-ij> 0$ then $f(x,y)\to0=f(0,0)$ as (x,y) tends to (0,0), hence, f(x,y) is continuous at (0,0).